Notes: `(y-1)sinx dx - dy = 0` Solve the first-order differential equation

Tuesday, 15 November 2016

$\displaystyle{\left({y}-{1}\right)}{\sin{{x}}}{\left.{d}{x}\right.}-{\left.{d}{y}\right.}={0}$ Solve the first-order differential equation

$\displaystyle{\left({y}-{1}\right)}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}-{\left.{d}{y}\right.}={0}$

To solve, express the equation in the form $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

So bringing same variables on one side, the equation becomes:

$\displaystyle{\left({y}-{1}\right)}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left.{d}{y}\right.}$

$\displaystyle{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}=\frac{{\left.{d}{y}}}{{{y}-{1}}}$

Then, take the integral of both sides.

$\displaystyle\int{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int\frac{{\left.{d}{y}}}{{{y}-{1}}}$

For the left side, apply the formula $\displaystyle\int{\sin{{\left({u}\right)}}}{d}{u}=-{\cos{{\left({u}\right)}}}+{C}$ .

And for the right side, apply the formula $\displaystyle\int\frac{{{d}{u}}}{{u}}\ldots$

$\displaystyle{\left({y}-{1}\right)}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}-{\left.{d}{y}\right.}={0}$

To solve, express the equation in the form $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

So bringing same variables on one side, the equation becomes:

$\displaystyle{\left({y}-{1}\right)}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left.{d}{y}\right.}$

$\displaystyle{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}=\frac{{\left.{d}{y}}}{{{y}-{1}}}$

Then, take the integral of both sides.

$\displaystyle\int{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int\frac{{\left.{d}{y}}}{{{y}-{1}}}$

For the left side, apply the formula $\displaystyle\int{\sin{{\left({u}\right)}}}{d}{u}=-{\cos{{\left({u}\right)}}}+{C}$ .

And for the right side, apply the formula $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$ .

$\displaystyle-{\cos{{\left({x}\right)}}}+{C}_{{1}}={\ln}{\left|{y}-{1}\right|}+{C}_{{2}}$

From here, isolate the y.

$\displaystyle-{\cos{{\left({x}\right)}}}+{C}_{{1}}-{C}_{{2}}={\ln}{\left|{y}-{1}\right|}$

Since C1 and C2 represent any number, express it as a single constant C.

$\displaystyle-{\cos{{\left({x}\right)}}}+{C}={\ln}{\left|{y}-{1}\right|}$

Then, eliminate the logarithm in the equation.

$\displaystyle{{e}}^{{-{\cos{{\left({x}\right)}}}+{C}}}={{e}}^{{{\ln}{\left|{y}-{1}\right|}}}$

$\displaystyle{{e}}^{{-{\cos{{\left({x}\right)}}}+{C}}}={\left|{y}-{1}\right|}$

$\displaystyle\pm{{e}}^{{-{\cos{{\left({x}\right)}}}+{C}}}={y}-{1}$

To simplify the left side, apply the exponent rule $\displaystyle{{a}}^{{m}}\cdot{{a}}^{{n}}={{a}}^{{{m}+{n}}}$ .

$\displaystyle\pm{{e}}^{{-{\cos{{\left({x}\right)}}}}}\cdot{{e}}^{{C}}={y}-{1}$

$\displaystyle\pm{{e}}^{{C}}\cdot{{e}}^{{-{\cos{{\left({x}\right)}}}}}={y}-{1}$

Since $\displaystyle\pm{{e}}^{{C}}$ is a constant, it can be replaced with C.

$\displaystyle{C}{{e}}^{{-{\cos{{\left({x}\right)}}}}}={y}-{1}$

$\displaystyle{C}{{e}}^{{-{\cos{{\left({x}\right)}}}}}+{1}={y}$

Therefore, the general solution is $\displaystyle{y}={C}{{e}}^{{-{\cos{{\left({x}\right)}}}}}+{1}$ .

Notes

Tuesday, 15 November 2016

$\displaystyle{\left({y}-{1}\right)}{\sin{{x}}}{\left.{d}{x}\right.}-{\left.{d}{y}\right.}={0}$ Solve the first-order differential equation

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Tuesday, 15 November 2016

Solve the first-order differential equation

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\left({y}-{1}\right)}{\sin{{x}}}{\left.{d}{x}\right.}-{\left.{d}{y}\right.}={0}$ Solve the first-order differential equation

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?